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a) Find \( \frac{d}{dx}(2\sin^{-1}(5x)) \) - HSC - SSCE Mathematics Extension 1 - Question 2 - 2005 - Paper 1
Question 2
a) Find \( \frac{d}{dx}(2\sin^{-1}(5x)) \). b) Use the binomial theorem to find the term independent of \( x \) in the expansion of \( \left( 2x - \frac{1}{x^2} \ri... show full transcript
Worked Solution & Example Answer:a) Find \( \frac{d}{dx}(2\sin^{-1}(5x)) \) - HSC - SSCE Mathematics Extension 1 - Question 2 - 2005 - Paper 1
Step 1
Find \( \frac{d}{dx}(2\sin^{-1}(5x)) \)
Answer
To find the derivative, we can use the chain rule. The derivative of ( \sin^{-1}(u) ) is ( \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} ).
Let ( u = 5x ). Thus, [ \frac{du}{dx} = 5 ] Now, applying the chain rule: [ \frac{d}{dx}(2\sin^{-1}(5x)) = 2 \cdot \frac{1}{\sqrt{1 - (5x)^2}} \cdot 5 = \frac{10}{\sqrt{1 - 25x^2}} ]
Step 2
Use the binomial theorem to find the term independent of \( x \) in the expansion of \( \left( 2x - \frac{1}{x^2} \right)^{12} \)
Answer
The binomial expansion of ( (a + b)^n ) is given by ( \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k ). Setting ( a = 2x ) and ( b = -\frac{1}{x^2} ), we need the power of ( x ) to equal zero: [ (2x)^{12 - k} \left(-\frac{1}{x^2}\right)^{k} \implies 2^{12 - k} (-1)^k x^{12 - k - 2k} = 2^{12 - k} (-1)^k x^{12 - 3k} ] Setting ( 12 - 3k = 0 ) gives ( k = 4 ). Thus, the term is: [ \binom{12}{4} (2x)^{8} \left(-\frac{1}{x^2}\right)^{4} ] Calculating this, we get: [\binom{12}{4} \cdot 2^8 \cdot (-1)^4 = 495 \cdot 256 = 126720]
Step 3
Differentiate \( e^{x}(\cos x - 3\sin x) \)
Answer
Using the product rule, we differentiate ( e^{x} ) and ( \cos x - 3\sin x ): [ \frac{d}{dx}[e^{x}( ext{function})] = e^{x} \cdot , \frac{d}{dx}( ext{function}) + (\text{function}) \cdot e^{x} ] Let the function be ( \cos x - 3\sin x ), then: [ \frac{d}{dx}(\cos x - 3\sin x) = -\sin x - 3\cos x\n] Thus, [ \frac{d}{dx}[e^{x}(\cos x - 3\sin x)] = e^{x}(-\sin x - 3\cos x) + e^{x}(\cos x - 3\sin x)\n] Combining like terms gives: [ e^{x}(-\sin x - 3\cos x + \cos x - 3\sin x) = e^{x}(-4\sin x - 2\cos x)\n]
Step 4
Hence, or otherwise, find \( \int e^{3x}\sin x \, dx \)
Answer
To find ( \int e^{3x}\sin x , dx ), we can use integration by parts or the method of undetermined coefficients. Assuming a solution of the form: [ \int e^{3x}\sin x , dx = A e^{3x} \sin x + B e^{3x} \cos x ] Differentiating and setting this equal to ( e^{3x}\sin x ) gives a system of equations to solve for constants ( A ) and ( B ). This process will lead to the integral result.
Step 5
Show that \( T = 3 + Ae^{-kt} \) satisfies this equation
Answer
Starting with the differential equation: [ \frac{dT}{dt} = -k(T - 3)\n] We can substitute the proposed solution format: [ T = 3 + Ae^{-kt}\n] Differentiating this, we get: [ \frac{dT}{dt} = -kAe^{-kt}\n] Substituting into the original equation, we check that: [ -kAe^{-kt} = -k(Ae^{-kt})\n] This holds true, thus confirming the solution.
Step 6
The temperature of the salad is 11°C after 10 minutes. Find the temperature of the salad after 15 minutes.
Answer
We know that when ( t = 10 ), ( T = 11 ): [ 11 = 3 + Ae^{-10k}\n] So, [ A = 8 + e^{10k}\n] To find the temperature at ( t = 15 ): [ T = 3 + A e^{-15k}\n] Using the previously computed value of ( A ), we can substitute and calculate the final temperature.